f(x)=root(4)(1+x) Use the binomial series to find the Maclaurin series for the function.

Binomial series is an example of an infinite series. When it is convergent at |x|lt1 , we may follow the sum of the binomial series as (1+x)^k where k is any number. The formula will be:
(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2) ...(k-n+1))/(n!) x^n
or
(1+x)^k = 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4+...
To evaluate the given function f(x) = root(4)(1+x) , we may apply radical property: root(n)(x) = x^(1/n) . The function becomes:
f(x)= (1+x)^(1/4)
or   f(x) =(1+x)^(0.25)
By comparing "(1+x)^k " with "(1+x)^(0.25) ”, we have the corresponding values:
x=x and k =0.25
Plug-in the values on the formula for binomial series, we get:
(1+x)^(0.25) =sum_(n=0)^oo (0.25(0.25-1)(0.25-2)...(0.25-n+1))/(n!)x^n
=sum_(n=0)^oo (0.25(-0.75)(-1.75)...(0.25-n+1))/(n!)x^n
=1 + 0.25x + (0.25(-0.75))/(2!) x^2 + (0.25(-0.75)(-1.75))/(3!)x^3 +(0.25(-0.75)(-1.75)(-2.75))/(4!)x^4+...
=1 + 0.25x + (-0.1875)/(1*2) x^2 + (0.328125)/(1*2*3)x^3+(-0.90234375)/(1*2*3*4)x^4+...
=1 + 0.25x -0.1875/2 x^2 + 0.328125/6x^3 -0.90234375/24x^4+...
=1 + x/4 -(3x^2)/32 +(7x^3)/128 -(77x^4)/2048 +...
Therefore, the Maclaurin series  for  the function f(x) =root(4)(1+x) can be expressed as:
root(4)(1+x)=1 + x/4 -(3x^2)/32 +(7x^3)/128 -(77x^4)/2048 +...